rs. 1760. - csquareajmer.comcsquareajmer.com/homebanner/timetable_n9ghxs.pdfrs. 32 1 1 13 5 42 = rs....
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1. Let the required sum be Rs. P. Then
Rs. P = Rs. 100 143 100 143 4 2
Rs.1 1 13 5
3 24 2
= Rs. 1760.
To find rate %
Since I = prt100
∴ r =100Ipt
2. Here P = Rs. 468.75, t = 2 5
1 or 3 3
years.
I = Rs. (500 − 468.75) = Rs. 31.25
∴rate p.c. = 100 31.25 3125 3
1005 46875 5
468.7453
= 4
3. Rs. 600 f or 2 years = Rs. 1200 for 1 years
And Rs. 150 for 4 years = Rs. 600 for 1 year
Int. = Rs. 90.
∴Rate = 90 1001800 1
= 5 %
To find Time
Since I = prt100
. ∴ t = 100I
Pr
4. Here interest = Rs. 15767.50 − Rs. 8500 = Rs. 7267.50
∴ t = 7267.50 1008500 4.5
= 19 years.
5. Let Principal = P, time = t ears, rate = t
Then P t t100 9
P
∴ t 2 =
100
9 ∴t =
10 13
9 3 ∴ rate =
13 %
3
Race Formula
. Rate = time = 1 10 1
100 3 %9 3 3
6. Let the annual payment be P rupees.
The amount of Rs. P in 4 years at 5%
The amount of Rs. P in 4 years at 5% = 100 4 5 120
100 100P P
“ “ “ 3 “ = 115P100
“ “ “ 2 “ = 110P100
“ “ “ 1 “ = 105P100
These four amounts together with the last annual payment of Rs. P will discharge the debt of
Rs. 770
∴115 P 110 P 105 P120 P
P100 100 100 100
= 770
∴550 P
100= 770 ∴
770×100
550= 140
Hence annual pay ment = Rs. 140
Theorem: The annual payment that will discharge a debt of Rs. A due in t years at the rate of interest r % per
annum is 100
1002
AA
rt t At
Proof: Let the annual payment be x rupees.
The amount of Rs. x in (t − 1) yrs at r % = 100 1
x100
t r
The amount of Rs. x in (t − 2) yrs at r % =100 2
x100
t r
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The amount of Rs. x in 2 yrs at r % = 100 2
x100
r
The amount of Rs. x in 1 yrs at r % = 100 1
x100
r
These (t −1) amounts together with the last annual pay ment of Rs. X will discharge the debt of Rs. A
∴100 1 100 2 100
x + x ... + x + x = A100 100 100
t r t r r
Or, x [{100 + (t − 1) r} + {100 + (t − 1) r} + ... {100 + r}]
= 100A
Or, x 1
100 1002
r t tA ∴x =
100
1100
2
A
r t tt
Note: 1 + 2 + 3 + ... + m = 1
2
m m
Using the above theorem:
Annual payment = 100 770 770 100
5505 4 55 100
2
= Rs. 140
7. By theorem payment = 848 100
4 3 44 100
2
= Rs. 200
8. Putting the values in the above formula:
80 = 100
5 4 55 100
2
A
Or, A = 80 550
100= Rs. 440
9. Let his deposit = Rs. 100
Interest for first 2 yrs. = Rs. 6
Interest for next 3 yrs. = Rs. 24
Interest for the last year. = Rs. 10
Total interest = Rs. 40
When interest is Rs. 40, deposited amount is Rs. 100
∴ when interest is Rs. 1520, deposited amount
= 100
152040
= Rs. 3800
Race formula:
Principal = + ...
1 1 2 2 3 3
1520 × 100Intereset ×100t r + t r + t r 2 × 3 + 3 × 8 + 1 × 10
= 1520 × 100
40= Rs. 3800
10. Let the sum it become Rs. 200.
∴ Interest = 200 − 100 = 100
Then, rate = 100 I
Pt=
100 ×100100×10
= 10%
Race formula:
Time × Rate = 100 (Multiple number of principal − 1)
Or, Rate = 100 × Multiple number of principal 1
time
Using the above formula: rate = 100 2 1
10= 10%
11. Rate = 100 3 1
20= 10%
Note: A generalized form can be shown as:
If a sum of money becomes ‘x’ times in‘t’ years as SI, the rate of interest rate is given by
100 x 1
%t
12. Using the above formula.
Time = 100 Multiple number of principal 1
Rate
= 100 4 1
5= 60 years.
13. Amount of 1st part =
1101st part
100
“ “ 2nd
part = 115
2nd part100
“ “ 3rd
part = 120
3rd part100
According to t he question, these amounts are equal
∴ 110 × 1st part = 115 × 2
nd part = 120 × 3
rd part
∴ 1sr part: 2nd
part: 3rd
par t=1 1 1
:110 115 120
:
Hence, dividing Rs. 2379 into three parts in the ratio 276: 264: 253, we have 1st part = Rs. 828,
2nd
part= Rs. 792, 3rd
part = Rs. 759.
14. P + SI for 3.5 yrs = Rs. 873
P + SI for 2 yrs = Rs. 756
On subtracting, SI for 1.5 yrs = Rs. 117
Therefore, SI f or 2 yrs = Rs. 117
21.5
= Rs. 156
∴ P = 756 − 156 = Rs. 600
And rate = 100 156600 2
= 13% per annum
15. Let the sum be Rs. x and the original rate be y% per annum. Then, new rate = (y + 3) % per annum.
∴3 2 2
100 100
x y x y= 300
xy + 3x − xy = 15000 or x = 5000
Thus, the sum = Rs. 5000
Race formula: Direct Formula
Sum = More Intereset × 100 300 100
2 3Time × More rate= 5000
16. Rate = 100(2 1) 100
7 7
100(4 121
100
7
Time years
OTHER METHOD: This question can be solved without writing anything.
Think like.
Doubles in 7 years
Trebles in 14 years
4 times in 21 years
5 times in 28 years
And so on.
17. Let the amount lent at 3% rate be Rs x, then
3% of x + 5% of (4000 – x) = 144
Or, 3x + 5 4000 – 5x = 14400
Or, 2x = 5600 X = 2800
Thus, the two amounts are Rs 2800 and (4000–2800) or Rs 1200
18. First rate of interest = 120 100
800 3
= 5%
New rate = 5 + 3 = 8%
New interest = 800 3 8
192.100
Rs
New amount = 800 + 192 = 992.
19. Simple interest for 5 years = Rs 300
Now, when principal is trebled, the simple interest for 5 years will also be treble the simple interest
On original principal for the same period. Thus SI for last 5 years when principal is trebled
=3 300 = Rs 900
Total SI for 10 yrs = 300 + 900 = Rs 1200
Theorem: A sum of Rs X is lent out in n parts in such a way that the interest on first part at r1% for t1
yrs, the interest on
Second part at r2% for t 2 yrs the interest on third part at r3% for t3 yrs. And so on, are equal, the ratio
in which the sum was divided in n parts is given by1 1 2 2 3 3
1 1 1 1: : : ............
n nr t r t r t r t.
Proof: Let the sum be divided into 1 2, ............ nS S S
1
2
3
int 100
1 1
int 100
2 2
int 100
3 3
int 100n
n n
Sr t
Sr t
Sr t
Sr t
[Since interest of all parts are equal]
1 2 3
int 100 int 100 int 100 int 100: : : ........ : : ................. :
1 1 2 2 3 3n
n n
S S S Sr t r t r t r t
= 1 1 2 2 3 3
1 1 1 1: : : ............
n nr t r t r t r t
20. Each Interest = 1 5 10 2 6 9
100 100
stpart ndpart
Or,
1 6 9 2727 : 25
2 5 10 25
26001 27 1350
27 25
2 2600 1350 1250
stpart
ndpart
stpart Rs
and ndpart Rs
21. Interest = Rs 840 – Rs 750 = 90
Time = 90 100
3750 4
yrs
Now, by the formula:
Sum = 100 100 575
500100 100 3 5
amountRs
rt
Note: There is a direct relationship between the principal and the amount
And is given by 100
100
AmountSum
rt
22. Use the formula
Principal = 100 100 2613 100 2613
2010100 100 30 130
AmountSum Rs
rt
Again by using the same formula:
100 30152010
100 5
100 3015,100 5
2010
1 100 3015 100 2010
5 2010
100 (3015 2010) 100 1005 100 100510
2010 5 2010 5 2010 5
t
or t
t
years
23. Let the sum be Rs X.
Interest = 8 4 32
100 100
x x
32 68
100 100
x xx
When interest is 68
100
x less, the sum is Rs X.
When interest is 340less, the sum is 100 340 50068
xRs
x
Direct formula: 100 100 340
340 500100 8 4 68
Sum Rs
24. We may consider that Rs (1800 – 1650) gives interest of Rs 30 at 4% per annum.
30 100
5 .150 4
Time yrs
25. After 2 yrs amount returned to Ramu
400 5 2
400 440100
Rs
Amount returned to Arun = 2% of Rs 440 = Rs 8.80.
26. Theorem: When different amounts mature to the same amount at simple rate of interest, the ratio of
the amounts
Invested are in inverse ratio of (100 + time rate). That is, the ratio in which the amounts are
invested is
1 1 2 2 3 3
1 1 1 1: : : .................. :
100 100 100 100 n nr t r t r t r t
Proof: We know that sum = 100
100
amount
rt
Let the sum invested be1 2, ............ nS S S ,
At the rate of 1 2 3, , .................. nr r r r for time
1 2 3, , , nt t t t yrs respectively Then
1 2 3: : : .................. : nS S S S
1 1 2 2 3 3
100 100 100 100: : : ....................... :
100 100 100 100 n n
A A A A
rt r t r t r t
[Since the amount (A) is the same for all]
1 1 2 2 3 3
1 1 1 1: : : ....................... :
100 100 100 100 n nr t r t r t r t
Soln: Therefore, the required ratio is this case is
1 1 1 1 1 1
: : : :100 2 5 100 3 5 100 4 5 110 115 120
27. Doubles in 4 yrs
3 times 4 2 = 8 yrs
4 times 4 3 = 12 yrs
8 times 4 7 = 28 yrs
Thus direct formula: X Times in = No. of yrs to double (X – 1)
8 times = 4(8 – 1) = 4 7 = 28 yrs.
28. Let the sum be Rs X, then
15 5 15 7144
100 200
,150 105 144 200
144 200640
45
x x
or x x
x Rs
Direct formula: Two equal amounts of money are deposited at r1% and r2% for t1 and t2 yrs
respectively. If the difference between
Their interest is di then sum = 1 1 2 2
100di
r t r t
Thus, in this case: sum = 144 100 144 100
64015 5 15 3.5 22.5
Rs
29.
1
1 1
2
2 2
1 2 1 2
500 210
100
500 210
100
10 10 2.5
rI r
rI r
I I r r
Or, r1–r2 = 2.5
10 = 0.25%
By Direct formula: When t1 = t2,
(R1–r2) =100 2.5 100
0.25%500 2
dI
Sum t
30. Let the sum be Rs X, then at 4% rate for 4 yrs the simple interest
= 4 4 4
100 25
x xRs
At 5% rate for 3 yrs the simple = 5 3 3
100 20
x xRs
Now, we have,
4 380
25 20
16 15, 80
100
8000
x x
x xor
X Rs
Quicker Method: For this type of question
Sum = 1 1 2 2
100 80 100
4 4 3 5
Difference
r t r t
= Rs 8000
31. Suppose the of interest = r% and the sum = Rs A
Now, A + 4
600;100
A r or, A + 600
25
rA
, 1 600 (1)
25
6, 650;
100
ror A
A rand A
, 1 600
25
6, 650; (2)
100
ror A
A rand A
Dividing (1) by (2), we have
1
600 (25 ) 2 1225 ; ,3 650 50 3 13
150
r
rOr
R r
Or, (50 + 2r) 13 = (50 + 3r) 12
Or, 650 + 26r = 600 + 3r; or, 10r = 50 r = 5%
Direct formula: If a sum amounts to Rs A1 in t1 years and Rs A2 in t2 years at simple rate of interest,
Then rate per annum =
2 1
1 2 2 1
100 A A
A t A t
In the above case,
R = 100 650 600 100 50
5%6 600 4 650 1000
32. Any sum that is paid back to the bank before the last instalment is deducted from the principal and not
from the interest.
Total interest = Interest on Rs 7000 for 3 yrs + Interest on (Rs 7000 – Rs 3000) = Rs 4000 for 2 yrs.
Or, (5450 + 3000 – 7000) = 7000 3 4000 2
100 100
r r
Or, 1450 = 210r + 80r 1450
290r = 5%
33. Suppose Rs X was lent at 6% per annum.
Thus, 6 5 (7000 ) 4 5
1600100 100
x x
3 7000
, 160010 5
x xor
3 14,000 2
, 160010
x xor
X = 16000 – 14000 = Rs 2000.
By Method of Alligation: Overall rate of Interest = 1600 100 32
%5 7000 7
Ratio of two amount = 2: 5
Amount lent at 6% = 7000
27
=Rs 2000
34. SI = 400 5 6
100
= Rs 120
35. Interest = Rs. 1 146 15 1
3064 365 4 100
= Rs. 1225 2 15 1
4 5 4 100
= Rs. 147
32= Rs. 4.59 nearly.
6%
32%
7
4%
10%
7
4%
7